Question

The value of $a^3+b^3+c^3-3abc$, if $a+b+c=15$ and $ab+bc+ca=74$ is?

• A. $12$
• B. $45$
• C. $42$
• D. $36$

Answer 1

The correct option is B $45$

Given :

$a+b+c=15$ and $ab+bc+ca=74$

Indentities:
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ (1)
$a^2+b^2+c^2=(a+b+c{)}^2-2(ab+bc+ca)$ (2)
Substituting (2) in (1):
$a^3+b^3+c^3-3abc=(a+b+c)\left(\right(a+b+c{)}^2-3(ab+bc+ca))$

Substituting the values from the question:

$a^3+b^3+c^3-3abc=\left(15\right)\left(\right(15{)}^2-3\left(74\right))$

$=\left(15\right)(225-222)$

$=\left(15\right)\left(3\right)$

$a^3+b^3+c^3-3abc=45$